-2x^2-4x+2 eching

$$-2 { x }^{ 2 } -4x+2$$

Omil

$$-2\left(x-\left(-\sqrt{2}-1\right)\right)\left(x-\left(\sqrt{2}-1\right)\right)$$

Kvadrat tenglamalaridan foydalanish qoidalari

Kvadrat koʻp tenglama bu orqali hisoblanadi: ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), bu yerda x_{1} va x_{2} ax^{2}+bx+c=0 kvadrat tenglamaning yechimlari.

$$-2x^{2}-4x+2=0$$

ax^{2}+bx+c=0 shaklidagi barcha tenglamalarni kvadrat formulasi bilan yechish mumkin: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Kvadrat formula ikki yechmni taqdim qiladi, biri ± qo'shish bo'lganda, va ikkinchisi ayiruv bo'lganda.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

-4 kvadratini chiqarish.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\left(-2\right)\times 2}}{2\left(-2\right)}$$

-4 ni -2 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+8\times 2}}{2\left(-2\right)}$$

8 ni 2 marotabaga ko'paytirish.

$$x=\frac{-\left(-4\right)±\sqrt{16+16}}{2\left(-2\right)}$$

16 ni 16 ga qo'shish.

$$x=\frac{-\left(-4\right)±\sqrt{32}}{2\left(-2\right)}$$

32 ning kvadrat ildizini chiqarish.

$$x=\frac{-\left(-4\right)±4\sqrt{2}}{2\left(-2\right)}$$

-4 ning teskarisi 4 ga teng.

$$x=\frac{4±4\sqrt{2}}{2\left(-2\right)}$$

2 ni -2 marotabaga ko'paytirish.

$$x=\frac{4±4\sqrt{2}}{-4}$$

x=\frac{4±4\sqrt{2}}{-4} tenglamasini yeching, bunda ± musbat. 4 ni 4\sqrt{2} ga qo'shish.

$$x=\frac{4\sqrt{2}+4}{-4}$$

4+4\sqrt{2} ni -4 ga bo'lish.

$$x=-\left(\sqrt{2}+1\right)$$

x=\frac{4±4\sqrt{2}}{-4} tenglamasini yeching, bunda ± manfiy. 4 dan 4\sqrt{2} ni ayirish.

$$x=\frac{4-4\sqrt{2}}{-4}$$

4-4\sqrt{2} ni -4 ga bo'lish.

$$x=\sqrt{2}-1$$

ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right) formulasi yordamida amalni hisoblang. x_{1} uchun -\left(1+\sqrt{2}\right) ga va x_{2} uchun -1+\sqrt{2} ga bo‘ling.

$$-2x^{2}-4x+2=-2\left(x-\left(-\left(\sqrt{2}+1\right)\right)\right)\left(x-\left(\sqrt{2}-1\right)\right)$$

Baholash

$$2-4x-2x^{2}$$